#!/usr/bin/env python
# -*- coding: utf-8 -*-
# 
# Copyright (c) 2017 Baidu.com, Inc. All Rights Reserved
# 

"""
File: run21.py
Author: zhangyang(zhangyang40@baidu.com)
Date: 2018/1/5 0005 15:32
"""
"""
题目描述
输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），
返回结果为复制后复杂链表的head。（
注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）
"""


class RandomListNode:
    def __init__(self, x):
        self.label = x
        self.next = None
        self.random = None


class Solution:
    # 返回 RandomListNode
    def Clone(self, pHead):
        if pHead is None:
            return None
        pCur = pHead
        # 复制next 如原来是A->B->C 变成A->A'->B->B'->C->C'
        while pCur is not None:
            node = RandomListNode(pCur.label)
            node.next = pCur.next
            pCur.next = node
            pCur = node.next
        pCur = pHead
        # 复制random pCur是原来链表的结点 pCur.next是复制pCur的结点
        while pCur is not None:
            if pCur.random is not None:
                pCur.next.random = pCur.random.next
            pCur = pCur.next.next
        head = pHead.next
        cur = head
        pCur = pHead
        # 拆分链表
        while pCur is not None:
            pCur.next = pCur.next.next
            if cur.next is not None:
                cur.next = cur.next.next
            cur = cur.next
            pCur = pCur.next

        return head
